How can I calculate whether a shelf is strong enough to support a falling cat? I'm building some shelves for my cat to climb on. 
The highest position the cat will be is 2 meters above the next shelf.
The cat has a mass of 6.5kg. The shelf has a maximum load of 20kg.
Using these parameters, how can I calculate whether the shelf's maximum load is sufficient to safely support the falling cat?
 A: If an object accelerates over a distance $d_1$ with a constant force $F_1$, then decelerates over a distance $d_2$ with a constant force $F_2$, then conservation of energy would give us that
$$F_1 d_1 = F_2 d_2$$
from which it would follow that
$$F_2 = F_1 \frac{d_1}{d_2}$$
In other words - if the cat is able to absorb the shock of his fall over a greater distance, the force will be less. Of course, when they have the chance, they do so beautifully by first stretching their paws, and then crouching down: this is nicely shown in this answer from which I reproduce the picture (originally by Etienne-Jules Marey):

Mathematically, if the total drop is 2 m and the cat can absorb the shock over a distance of 20 cm (roughly how far the center of mass drops from first contact to stationary cat) then the average force during the landing is ten times its weight.
Note that the "rating" of a shelf assumes the ability to "plop down" the rated weight - if you took a 20 kg book and placed it on the shelf that was rated for 20 kg, you would not expect the shelf to break (although you will briefly exert a force much greater than 200 N). This shows that you need to make a distinction between static and dynamic limit: for things like shelves, they usually specify the load limit as "the biggest object you should put on this shelf",  knowing full well that when you put it down roughly, the instantaneous force will be much greater.
If found an interesting reference from the Composite Panel Association - they give design rules for shelves, including detailed calculations of local stress and rupture strength of particle board. Without knowing the details of your shelf setup I can't comment on the applicability of these calculations - but you might find there is some useful information there. One thing to note, for instance, is that the stress on the shelf depends on the distance between the supports as well as how the edges are supported. If you are worried, you can create a center support for your shelf and increase the load capacity by more than 2x. Note also that the design rules they give are for keeping the deflection of the shelves below a small amount (0.1 inch for 24 inch shelf, or about 0.4% of the span). Rupture will occur at much greater deflections.
I think your cat will be quite safe. But you might give some lessons in etiquette to his fellow cats. Push him off the top shelf? Really?
A: When the cat leaves the upper shelf, it will be accelerated by the Earth's gravitational field toward the lower shelf at a rate of 9.8 meters per second, per second (9.8 m/sec^2).  Assuming its initial velocity was zero, at the time the cat lands on the lower shelf its velocity will be 19.6 m/sec.
The cat's momentum at the lower shelf will be 6.5 Kg * 19.6m/sec = 127.4 Kg m/s.
Let's assume that the cat's legs are 15 centimeters fully extended.  When the cat lands, it's legs will fold up under it, reducing their effective length to 5 centimeters.  The cat will decelerate from 19.6 m/sec to zero while collapsing 10 centimeters.  It's average velocity during deceleration will be 9.8 m/sec, so it will take about 0.1 seconds to come to a complete standstill on the lower shelf, and its deceleration will be 19.6 / 0.1 = 196 m/sec^2.
The impulse-momentum theorem says that the impulse the shelf applies to the cat will equal the cat's change in momentum.  As the momentum drops from 127.4 Kg m/s to zero, the impulse will be 127.4 newton seconds.
We can derive force from the impulse-momentum equation:
Force * t = m * ∆v
Force * 0.1 sec = 6.5 Kg * -19.6 m/sec
Force = -127.4 Kg m/sec / 0.1 = -1,274 newtons
Let's check the force derived from the impulse-momentum theorem, against the force we can derive from Newton's 2nd law of motion:  Force = 6.5Kg * -196m/sec^2 = -1,274 newtons.
Newton's 3rd law of motion says that the force applied to the cat, will be applied by the cat to the lower shelf.
Therefore, the shelf will bear a force of 1,274 newtons.  As the shelf was designed to carry 20 Kg * 9.8 = 196 newtons, the shelf will not withstand the cat's impact unless the cat, through superior muscle tone and reactions, extends its deceleration time.
You need either to place a thick foam pad on the shelf to lengthen the deceleration time, or to install a much stronger shelf.
Caution:  My assumption about the cat's deceleration may be incorrect.
A: You will have to estimate how quickly the cat goes from full speed to completely stopped.  Then, given the cat's initial velocity as it gets to the shelf, use the impulse-momentum theorem.  The equation is:
$$F = \frac {\Delta p}{\Delta t}$$
Is that enough info, or do you need an additional equation to calculate the initial velocity of the cat?
